Differentials, pseudo-differentials, and locality

Main Question or Discussion Point

If [itex]\psi(x,0)[/itex] has a compact support, and
[tex]
i\partial_t\psi(x,t) = -\partial_x^2 \psi(x,t),
[/tex]
then [itex]\psi(x,t)[/itex] does not have a compact support for any [itex]t>0[/itex].

Claim 2:

If [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are the same in some environment of a point [itex]x_0[/itex], then
[tex]
\partial_x^2 \psi_1(x_0) = \partial_x^2 \psi_2(x_0)
[/tex]

Claim 3:

If [itex]\psi(x,0)[/itex] has a compact support, and
[tex]
i\partial_t\psi(x,t) = \sqrt{1 - \partial_x^2} \psi(x,t),
[/tex]
then [itex]\psi(x,t)[/itex] does not have a compact support for any [itex]t>0[/itex].

Question:

If [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are the same in some environment of [itex]x_0[/itex], will it follow that
[tex]
\sqrt{1 - \partial_x^2}\psi_1(x_0) = \sqrt{1 - \partial_x^2}\psi_2(x_0)?
[/tex]

Thoughts:

According to my understanding, at least with some assumptions, the claims 1,2,3 are all true. If I had not mentioned the claims 1 and 2, some people might have answered to my question, that the pseudo-differential operator [itex]\sqrt{1 - \partial_x^2}[/itex] does not possess the locality property I'm asking in the question, because we know that it is non-local in the sense of the claim 3. However, we know that the ordinary differential operator [itex]\partial_x^2[/itex] has the non-locality property in the sense of claim 1 too, and still it has the locality property in the sense of claim 2. So without better knowledge, it could be that the pseudo-differential operator has this locality property too. But what's the truth?

I think that the answer to my question is "no", and the instantaneous spreading caused by [tex]\sqrt{1 - \partial_x^2}[/tex] is significantly faster than the instantaneous spreading caused by [tex]\partial_x^2[/tex].

I just realized that I don't yet know how to prove that the non-relativistic Schrödinger equation causes instantaneous spreading. Here's the problem:

If we want to satisfy the heat equation

[tex]
\partial_t u(t,x) = \partial_x^2 u(t,x)
[/tex]

with some initial condition [itex]u(0,x)\geq 0[/itex] which has compact support, we can accomplish it with a formula

I know some physicists like to say that "well look at the propagator. It is non-zero for arbitrarily large intervals, and hence the instantaneous spreading." You might as well try to insist that Fourier transforms (or inverse transforms) will never give functions with compact support, because the plane waves extend to infinities. Of course canceling can occur!

An interesting remark about the heat equation is that if we fix some [itex]x'[/itex] outside the original compact support, then the Taylor series

is not analytic because of the branch cuts at [itex]p=\pm i[/itex]. Hence [itex]\partial_t\psi(0,x)[/itex] does not have a compact support, and the instantaneous spreading is faster than with the heat equation.